Step 1 (Formula / Governing Law): (1 Mark)

The equation of the line is given by:

$$\frac{x-4}{-2} = \frac{y-0}{6} = \frac{z-1}{-3} = \lambda$$

Any general point $P$ on the line is given by $(x, y, z) = (-2\lambda + 4, 6\lambda, -3\lambda + 1)$. The direction ratios of the line are $\vec{b} = (-2, 6, -3)$. If $A$ is the point $(2, 3, -8)$, the vector $\vec{AP}$ must be perpendicular to the line, i.e., $\vec{AP} \cdot \vec{b} = 0$.

Step 2 (Step-by-Step Working): (3 Marks)

The vector $\vec{AP} = (-2\lambda + 4 - 2, 6\lambda - 3, -3\lambda + 1 - (-8)) = (-2\lambda + 2, 6\lambda - 3, -3\lambda + 9)$.

Since $\vec{AP} \perp \vec{b}$:

$$-2(-2\lambda + 2) + 6(6\lambda - 3) - 3(-3\lambda + 9) = 0$$ $$4\lambda - 4 + 36\lambda - 18 + 9\lambda - 27 = 0$$ $$49\lambda - 49 = 0 \implies \lambda = 1$$

Substituting $\lambda = 1$ into the general point $P$:

$$x = -2(1) + 4 = 2$$ $$y = 6(1) = 6$$ $$z = -3(1) + 1 = -2$$

The foot of the perpendicular is $(2, 6, -2)$.

Now, calculate the perpendicular distance $AP$:

$$AP = \sqrt{(2-2)^2 + (6-3)^2 + (-2 - (-8))^2}$$ $$AP = \sqrt{0^2 + 3^2 + 6^2} = \sqrt{9 + 36} = \sqrt{45} = 3\sqrt{5}$$

Step 3 (Final Answer): (1 Mark)

• Foot of the perpendicular: $\boxed{(2, 6, -2)}$
• Perpendicular distance: $\boxed{3\sqrt{5} \text{ units}}$